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Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

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Given : A rhombus ABCD whose diagonals intersect each other at O. 

To prove : A circle with AB as diameter passes through O. 

Proof : ∠AOB = 90° 

[Diagonals of a rhombus bisect each other at 90°] 

⇒ ∆AOB is a right triangle right angled at O. 

⇒ AB is the hypotenuse of A B right ∆AOB. 

⇒ If we draw a circle with AB as diameter, then it 

will pass through O. because angle is a semicircle is 90° and ∠AOB = 90° Proved.

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